岛屿数量（力扣）

**********题目链接
BFS解法 ？？？时间超限？？？

class Solution {
public:
    int numIslands( vector<vector<char>>& grid ) {

        int m = grid.size();
        int n = grid[0].size();

        if( n == 0 || m == 0 ) {
            return 0;
        }

        int count = 0;

        for( int i = 0; i < m; i++ ) {
            for( int j = 0; j < n; j++ ) {
                if( grid[i][j] == '1' ) {
                    count++;
                    Search( grid, i, j, m, n );
                }
            }
        }

        return count;


    }

    void Search( vector<vector<char>>& grid, int i, int j, int m, int n ) {
        queue<int>q;
        q.push( i * n + j );

        while( !q.empty() ) {
            int front = q.front();
            q.pop();
            int x = front / n;
            int y = front % n;
            grid[x][y] = '0';

            //up
            if( x - 1 >= 0  && grid[x - 1][y] == '1' ) {
                q.push( ( x - 1 )*n + y );
            }

            //down
            if( x + 1 < m  && grid[x + 1][y] == '1' ) {
                q.push( ( x + 1 )*n + y );
            }

            //left
            if( y - 1 >= 0  && grid[x][y - 1] == '1' ) {
                q.push( x * n + y - 1 );
            }

            //right
            if( y + 1 < n  && grid[x][y + 1] == '1' ) {
                q.push( x * n + y + 1 );
            }
        }
    }
};

DFS解法 这个过了

class Solution {
public:
    int numIslands( vector<vector<char>>& grid ) {
        int m = grid.size();
        int n = grid[0].size();
        int count = 0;

        for( int i = 0; i < m; i++ ) {
            for( int j = 0; j < n; j++ ) {
                if( grid[i][j] == '1' ) {
                    count++;
                    Search( grid, i, j, m, n );
                }
            }
        }

        return count;


    }

    void Search( vector<vector<char>>& grid, int i, int j, int m, int n ) {
        grid[i][j] = '0';

        //up
        if( ( i - 1 >= 0 ) && grid[i - 1][j] == '1' ) {
            Search( grid, i - 1, j, m, n );
        }

        //down
        if( ( i + 1 < m ) && grid[i + 1][j] == '1' ) {
            Search( grid, i + 1, j, m, n );
        }

        //left
        if( ( j - 1 >= 0 ) && grid[i][j - 1] == '1' ) {
            Search( grid, i, j - 1, m, n );
        }

        //right
        if( ( j + 1 < n ) && grid[i][j + 1] == '1' ) {
            Search( grid, i, j + 1, m, n );
        }
    }
};